help-39

[barren pawn]

I need help with this limit i cant figure out how to start solving this the awnser should end up being e^-2

[barren pawn]

I get to this point znd get stuck

<@&286206848099549185>

[barren pawn]

?

<@&286206848099549185>

Is this allowed?

[pearl pond]

@barren pawn Has your question been resolved?

[pearl pond]

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[woeful bloom]

what does with respect to T mean?

[plush bramble]

do you know partial derivatives

[woeful bloom]

yea

[plush bramble]

it means find $\frac{\pa \rho}{\pa T}$

[jolly parrot]

riemann

[woeful bloom]

but what equation do I differentiate?

I only know the rate of change one which is d(A) - d(B) divided by the magnitude of AB

[plush bramble]

you approximate this using your graph

https://www.youtube.com/watch?v=f_Q0qlwlclQ has the average rate of change formulas

[woeful bloom]

yeah that's the one I used in the thumbnail

wait no I had the magnitude at the bottom

alright, i'll look into it. Thanks!

.close

[pearl pond]

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[last summit]

I'm hoping I could get a small hint on this problem

I've tried from the definition of differentiability, that since f is differentiable we can write, f(x+h)=f(x)+f'(x)h+E(h) and similarly with a substitution on the f(x-h)

[jolly parrot]

Austin

[last summit]

Which doesn't really seem relevant/useful

So I think I am supposed to use another strategy

[cursive wraith]

have you tried f(x+h) = ... + E(h^2)?

using twice differentiability?

[last summit]

No I don't know that definition

[cursive wraith]

ok

[last summit]

I wasn't sure how to rewrite in terms of the double limit honestly

[cursive wraith]

f' is differentiable, so f'(x+h) = ...

[last summit]

=f'(x)+f''(x)h+E(h)

[cursive wraith]

yes

so integrate this between x and x+h

[last summit]

I can't use integrals

but I'm not sure what you mean by that aswell

[cursive wraith]

ok sure

let's do this otherwise

f''(x)h^2 = ...

[last summit]

-E(h)-f'(x)

[cursive wraith]

h^2

[last summit]

ah

multiplied by h

then

[cursive wraith]

and missing something else

[last summit]

f'(x+h)=f'(x)+f''(x)h+E(h)
f'(x+h)h-f'(x)h-E(h)h=f''(x)h^2

[cursive wraith]

-E(h)h = E(h^2)

f''(x)h^2 = f'(x+h)h-f'(x)h+E(h^2)

now, hf'(x) = ...
hf'(x+h) = ...

[last summit]

-E(h)h=E(h^2) ?

[cursive wraith]

Yes

[last summit]

why is that

[cursive wraith]

-E(...) = E(...)

[last summit]

sure

but E(h)h=E(h^2)?

[cursive wraith]

xE(...) = E(x...)

[last summit]

why would that be true

[cursive wraith]

Definition of E

[last summit]

E(h)/h ->0 as h->0

is definition of E(h)

right

[cursive wraith]

g(h) = E(f(h)) if g(h)/f(h) -> 0

if g(h) = hE(h)

Then g(h)/h = E(h)

so (g(h)/h)/h -> 0

so g(h)/h^2 -> 0

So g(h) = E(h^2)

and vice versa

[last summit]

okay forgive me if I need some time to reason through all of that

one sec

[last summit]

here, this is as h->0 yes?

[cursive wraith]

yes

all limits are when h -> 0 here

[last summit]

Okay I see

I am going to try to solve with your hints

ty

[last summit]

Okay so I can do these steps

but I'm still not seeing the algebra required to get it into the form I need

right now I have

$$f''(x)=\frac{f'(x+h)h-f'(x)h+E(h^2)}{h^2}$$

[jolly parrot]

Austin

[last summit]

and I have replacements for f'(x+h)h and for f'(x)h

but like I've tried in multiple ways to substitute

and just nothing nice comes of it

what do you recommend?

[cursive wraith]

f(x+h) - f(x) = ...

and f(x+2h) - f(x+h) = ...

[last summit]

f(x+h)-f(x)=hf'(x)+E(h)

f(x+2h)-f(x+h) this I am unsure of

the 2h is throwing me off with what you want me to rewrite it as

[cursive wraith]

hehe

it usually throws off everyone a bit

[last summit]

f(x)+2hf'(x)+E(2h)-f(x)-h'f(x)-E(h)

hf'(x)+E(2h)-E(h) ?

hf'(x) + E(h)

is that all?

[cursive wraith]

this is the part where it's a bit obnoxious

so in order to cut through this mess

[cursive wraith]

we'll ignore this

and instead

[last summit]

obnoxious indeed

[cursive wraith]

suppose h is fixed

and s->0

f(x+h+s) = ...

[last summit]

f(x+h)+f'(x+h)s+E(s) ?

[cursive wraith]

ooh yeah

now h = s 😎

[last summit]

haha what

why can we do that

f(x+2h) = f(x+h)+f'(x+h)h+E(h) it becomes

[cursive wraith]

there is a very rigorous proof behind that that I will not go through

[cursive wraith]

you got it

[last summit]

hm

where do we even have a f(x+2h)

[cursive wraith]

try to replace everything in the sum here

[last summit]

$$f''(x)=\frac{f'(x+h)h-f(x+h)+f(x)+E(h)+E(h^2)}{h^2}$$

[jolly parrot]

Austin

[cursive wraith]

hE(h), not E(h)

[last summit]

why?

f'(x)h=f(x+h)-f(x)-E(h)

[cursive wraith]

wait sorry

so this won't be enough what

[last summit]

yeah when the E(h) terms got introduced back in that is where I was getting stuck when trying on my own

[cursive wraith]

there is only one thing left to do

[last summit]

which is

[cursive wraith]

start back from the original expression

$\frac{f(x+h)-2f(x)+f(x-h)}{h^2} = \frac{\frac{f(x+h)-f(x)}{h}-\frac{f(x)-f(x-h)}{h}}{h}$

[jolly parrot]

rafilou2003

[cursive wraith]

and this will be tough to continue but I'm trying to remember

[last summit]

that's a tricky simplification

[cursive wraith]

wait a sec

you're allowed to use E notation

but you're not allowed to use its result on integration of E notation

[last summit]

yeah we have not covered any integration

E(h) just means that E(h)/h -> 0 as h->0

[cursive wraith]

What about taylor-young theorem?

[last summit]

I'm not sure what that is, but we have covered taylor's theorem as in taylor's series

[cursive wraith]

yesyesyes

[last summit]

if that is the same thing

ah I was hoping you wouldn't say that

XD

Taylor's theorem is such a mess

[cursive wraith]

do you have the result $f(x+h) = f(x) + hf'(x) + \frac{h^2}{2}f"(x) + E(h^2)$

[jolly parrot]

rafilou2003

[last summit]

I likely do

I think I will try this out with a taylors theorem approach on my own for a bit

and let you know if I get stuck later?

if that is fine

[cursive wraith]

PLEASE use this it's so much simpler

yes ping me for more help

[last summit]

Okay awesome tysm

[pearl pond]

@last summit Has your question been resolved?

[cursive wraith]

Final hint because I gtg to sleep :(
||Using the formula I gave you above, f(x+h) = ...||
||Then f(x-h) = || ||f(x) -hf'(x) + (-h)²/2 f"(x) + E((-h)²)||
||Then f(x+h) - 2f(x) + f(x-h) = || ||h²/2 f"(x) + h²/2 f"(x) + E(h²)||

[last summit]

That is alright, Ty very much for your help. I’ve just been brushing up on Taylor’s series

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[hard crystal]

Find no. Of terms in expansion of
(1+x+x⁴)^n

[pearl pond]

@hard crystal Has your question been resolved?

[hard crystal]

I just want my answer to be checked

[warm ledge]

Well what's your answer?

[hard crystal]

4n-2, n>=2@warm ledge

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@hard crystal Has your question been resolved?

[pearl pond]

@hard crystal Has your question been resolved?

[pearl pond]

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[finite crown]

https://gyazo.com/3e6357ee7bc6e00a5c3fce92970ae1eb

[finite crown]

ok

ive got a which is 2

just the radius

b is easy too az z >= 1 so we can equation 1 = acos(u) u = 1/3pi

but what is c

how does one work out C

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[midnight haven]

Question

[midnight haven]

Can [
\m\P{A\given B} = \m\P{A}
]
imply that $B$ is equal to the sample space $S$?

[vestal tapir]

imply?

[toxic lichen]

wdym by "can <...> imply" as opposed to "does <...> imply"?

[midnight haven]

i meant the latter

[toxic lichen]

no, P(A|B) = P(A) does not imply B = S.

it only means A is independent of B.

[cursive wraith]

only implies that A and B are independent

[midnight haven]

but why wouldn't it be okay [
\m\P{A\given S}= \f{\m\P{A\cap S}}{\m\P S} = \f{\m\P A}1 = \m\P A
]

[midnight haven]

or am i missing something here

[vestal tapir]

it's a necessary, not sufficient

it's the opposite of what imply means

[timid spindle]

Yeah thats showing the other implication direction, not the one you asked

[toxic lichen]

hol up

[midnight haven]

ah i see my mistake

okay

[toxic lichen]

how are you getting P(A&S)/P(S) = P(A)/1

[midnight haven]

its this implication in reverse

[midnight haven]

isn't the intersection of a set with its sample space equal to itself?

and like

the probability of the sample space occurin is 1

[timid spindle]

Yeah its fine

[toxic lichen]

ok no yeah it is

[midnight haven]

also just for my information; how would you explain the meaning of the formula [
\m\P{A\given B} = \f{\m\P{A\cap B}}{\m\P B}
]

[midnight haven]

its just for my own input and i heard a lot of explanations

[vestal tapir]

it's the definition

[timid spindle]

Venn diagram

[sharp laurel]

obligatory 3b1b has a video on it

but yeah venn diagram

[sharp laurel]

try to see what happens when A and B are independant

give an example

[midnight haven]

kinda bad that way tho, no?

because A|B itself is not an event

[sharp laurel]

no

its intuitive

think that the |B means youre only looking at the space where B happened

[midnight haven]

right yes

[sharp laurel]

so P(A|B) = P(A and B) / [normalization = P(B)]

[midnight haven]

thats the thing though

[sharp laurel]

its P(B) because its like the probablity that we are indeed in the space wehre B happened

[midnight haven]

Sure but like

assume the probabilities in the sample space are like uniformly distributed, then [
\m\P{A} = \f{|A|}{|S|}
]

[midnight haven]

at least thats how i think of it

but in your case, we are considering P(B) as like the sample space in the probability?

its just a bit weird for some reason to me

[sharp laurel]

i guess yeah

[timid spindle]

pretty much you're changing the sample space to B

[sharp laurel]

whats weird about it

[midnight haven]

why dont we consider |B| instead of P(B)?

[sharp laurel]

because P(B) is sorta |B|

if its evenly distributed as you said P(A|B) = |A+B|/|S| * 1/(|B|/|S|) = |A+B|/|B|

in a more advanced way P(B) is pretty much measuring B in the probabliy space giving you its "size"

[midnight haven]

oh i think i get it now yeah okay

[midnight haven]

is [
\m\mu{A\given B} = \f{\m\mu{A\cap B}}{\m\mu B}
]
axiomatic in that sense?

[sharp laurel]

by mu you mean measure?

[midnight haven]

yes

[sharp laurel]

i dont know measure theory like at all lol

srry

just how it formalizes probability in the sense P(A) = measure of A in some way

[timid spindle]

The RHS is definitely something you can define for any measure (μ(B)=/=0)

Whether you'd call that mu of "A given B" outside the context of probability i do not know

[midnight haven]

i see okay

okay one last question

[rustic gate]

youre here

[midnight haven]

hi yes

so like @rustic gate

[rustic gate]

the RHS defines a probability measure whenever \mu(B) is finite and non-zero

thats the whole point of dividing by P(B)

[midnight haven]

[
A_B = \m{\bigcup_{i\ge 1}}{\bigcap_{j<i} \comp{(B_j)}, A_i B_i}
]
Where $A_i$ and $B_i$ are elements of $A$ and $B$

[rustic gate]

holy

[sharp laurel]

what the fuck?

[rustic gate]

what are you trying to do

[rustic gate]

💀

[cursive wraith]

AiBi ?

[midnight haven]

whats the intuition behind this? This is the definition of conditional probability as the probability of a conditional event A_B

[rustic gate]

?

[cursive wraith]

I can assure you, this is NOT a probability

[rustic gate]

your notation is still nonsense

[cursive wraith]

You might try to define an event but it's really all over the place

[midnight haven]

oh no this is an event yes, but like apparently it can be shown that [
\m\P{A_B} =\f{\m\P{A\cap B}}{\m\P B}
]

[rustic gate]

no like

your notation doesnt make sense

where are you pulling this from

[midnight haven]

idk i pulled it online from a site explaining conditional probabilities and i was confused

its apparently called

"the Goodman–Nguyen–Van Fraassen conditional event"

whatever that means

[sharp laurel]

probably not the best site for beginners

or well

people trying to be explained conditional probablity

[cursive wraith]

Wtf it's written exactly like that on wikipedia

[timid spindle]

Snow was getting at that you've defined a new measure ν defined as ν(Α) = μ(Α & B)/μ(B)

[midnight haven]

yeah the site probably copied it from there

[timid spindle]

For some measure μ, ν is a prob measure when μ(B) finite and non zero

[rustic gate]

where the hell are you getting this

i dont see that formula on wikipedia

[midnight haven]

wait can you share the link im curious to see if there is a better explanation on there

[midnight haven]

yeah i think i got that part

[cursive wraith]

Nope, but here's the "reference" wikipedia refers to

https://www.sciencedirect.com/science/article/pii/S000437022030103X

[rustic gate]

where did you find this formula

[midnight haven]

okay i just found it in the wiki too lol

also

wtf is this

[midnight haven]

let me link it wait

https://en.wikipedia.org/wiki/Conditional_probability

you can scroll down a bit

to the definitions

also i guess i got my question answered 💀 the above was just random curiosity

thanks everyon

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[pulsar fractal]

Solve the system of equations with the following extended coefficient matrix and describe it Solution quantity geometric

Sorry If translation is bad

I do not understand how this Is supposed to be solved, as we have 4 unknowns but only two equations

How would I even start?

<@&286206848099549185>

[pearl pond]

@pulsar fractal Has your question been resolved?

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[pulsar fractal]

.reopen

[pearl pond]

✅

[pulsar fractal]

<@&286206848099549185>

[feral leaf]

Do you know gaussian elimination/rref?

[pearl pond]

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[pearl pond]

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[flat shale]

my teacher was giving a lecture and she came to this slide. with the two examples.

[flat shale]

can someone help me with the first equation with that rate of change formula x and x+h?

i have no idea what she means

@calm wing

[calm wing]

uh

[calm wing]

same

what's h

[flat shale]

i have no idea

can anyone else help

[lunar apex]

Definition of derivative is

lim h —> 0 = (f(x+h) -f(x))/h

so it may be asking take the derivative of the equation using definition and plugging in 0 for h?

[flat shale]

idk

[pulsar fractal]

brb in 10 mins btw

[pearl pond]

@flat shale Has your question been resolved?

[pearl pond]

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[ember talon]

Are these the same?

[plush bramble]

yes

the repeated index is a dummy variable

[ember talon]

Cool, thanks. Just wanted to make sure

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[midnight haven]

Hi, I'm trying to solve problem (b) of this, can someone please help me? Thanks

[midnight haven]

Got this, so far

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[flat kayak]

not sure if im making a mistake or what

(sendingh

[merry carbon]

What'd you do here?

[flat kayak]

IDK

[merry carbon]

Also comment: could have changed ln(x^3) to 3ln(x)

[flat kayak]

let me do this and retry

[merry carbon]

it would make your life a bit easier (but you didn't differentiate x^3 properly in your chain rule there either)

[merry carbon]

• sus common denominator here

[flat kayak]

idk whats going on

judging from this my answer already doesn't look like any of the answer choices

all of them have a radical in the numerator

[robust sentinel]

you can multiply to make it one whole fraction on line 2

[merry carbon]

Well you can multiply numerator and denominator by "something" that'll make your life much easier to work with

[flat kayak]

can i combine the sqrt xs to make 3sqrtx

or is that illegal

[flat kayak]

what is it ☹️

[merry carbon]

Look at what some of your options have in the denominator

And what you have here

[flat kayak]

i see the 1 + sqrtx

but wat do i do with it

😭

[merry carbon]

It would be nice to keep it in the denominator of that term

But that 2sqrt{x} is annoying

We could change it slightly by multiplying top and bottom to make it a bit nicer

[flat kayak]

so we would get um

4x on the bottom??

then i would assume something simplifies in the numberator to make it 2x

idk let me try

$\frac{2\sqrt{x}}{4x(1+\sqrt{x}}-\frac{3}{x}

pls work

[merry carbon]

You need a $ at the end

[flat kayak]

$\frac{2\sqrt{x}}{4x(1+\sqrt{x}}-\frac{3}{x}$

[jolly parrot]

jacks

[flat kayak]

i have this now

[merry carbon]

Yep \catthink you could have that as $\frac{\sqrt{x}}{2x(1 + \sqrt{x})} - \frac{3}{x}$ to be fair too, a bit nicer

[jolly parrot]

@merry carbon

[merry carbon]

(cancelling down the common 2's)

[merry carbon]

Now try and common denominator this here

[flat kayak]

uhhhh

ok so it's missing a 2 and the 1+sqrtx

so would i doooo

$\frac{3}{x}\frac{2(1+\sqrt{x}}{2(1+\sqrt{x}}$

[jolly parrot]

jacks

[flat kayak]

you get what i mean

[merry carbon]

I do, but yep, turn 3/x into that and then you can easily common denominator it

[flat kayak]

kk

i got c

[merry carbon]

You might have simplified wrong

[flat kayak]

fawl

fawk

this is my work

[flat kayak]

wtf how did i get positive 6

yea i was just looking at it

😭

omfg

[merry carbon]

it happens sometimes

[flat kayak]

@merry carbon TYSM

you're a life saver bruh

i am free.

[merry carbon]

definitely free for sure

Have a good one

[flat kayak]

you too

🫶

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[signal wren]

For this one, I don't know where I am going wrong in my calculations.

[pearl pond]

@signal wren Has your question been resolved?

[pearl pond]

@signal wren Has your question been resolved?

[pearl pond]

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[stable cove]

what do i after i do the antiderivatives

[primal orchid]

!show

[pearl pond]

Show your work, and if possible, explain where you are stuck.

[primal orchid]

If you mean, how do you find constants after doing an integration... Use the initial conditions

[stable cove]

what do u mean by inital conditions

because Ik that I have to do the antiderivatives

but idk whats the y=0 and t= 0 for

[primal orchid]

y(0)=0

To determine the constants left over after integration

After one integration, you have 1 constant

[stable cove]

so u suppose to do 0 = x -1/pi^pit

[primal orchid]

You use the initial equation to solve for the constant

[stable cove]

oh ok

can u check my answer after i am done

[primal orchid]

I can try, if you send all your work

[primal orchid]

where did x come from?

[stable cove]

the antideriviates

[primal orchid]

x is not the independent variable in that question

I guess you mean x is the constant

But that notation will make people upset

[stable cove]

no the one in the equation

arent u doing antiderivates of the equation first then find the constant

[primal orchid]

You're saying that the integral of 1 is x?

But we are not integrating with respect to x

We're integrating over t

[stable cove]

oh is t

nvm

is this right

[primal orchid]

No, because the first equation is not true

Maybe you mean the integral of lhs is equal to rhs?

But the notation is so bad that you would get points off

[stable cove]

so how would i do it right

i don't get it

[primal orchid]

Have you done integrals before?

[stable cove]

yes but i forgot

[primal orchid]

https://math.libretexts.org/Courses/Monroe_Community_College/MTH_211_Calculus_II/Chapter_5%3A_Integration/5.2%3A_The_Definite_Integral

This should refresh your memory

[stable cove]

oh wait nvm i don't know how to do this one

[stable cove]

do u have like a question or a sample that is is similar to the question

[stable cove]

is this right

[pearl pond]

@stable cove Has your question been resolved?

[stable cove]

<@&286206848099549185>

?

[pearl pond]

@stable cove Has your question been resolved?

[pearl pond]

@stable cove Has your question been resolved?

[pearl pond]

@stable cove Has your question been resolved?

[strange cargo]

bot gave up asking every hour

now asks every second

is it + - (1/pi)e^pi+?

or t instead of +?

[pearl pond]

@stable cove Has your question been resolved?

[pearl pond]

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[pearl pond]

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[teal cedar]

how does he go from that to that?

[shrewd basin]

He doesn’t go from that to that directly

[teal cedar]

okay

but how

[shrewd basin]

Like he said long division

And attempt to use Euclidean algorithm

[teal cedar]

okay, what did he do exactly?

i tried using euclidean algorithm

but it only grows bigger and bigger

im still not sure how he got to that result

[shrewd basin]

have you tried doing long division

[teal cedar]

no, because i dont know how to do that with gcd

like, i know what it is

but i dont know how to apply it in this situation

im sure its dumb as hell

but i just dont know

[shrewd basin]

Youre not doing that with gcd you’re doing that with two polynomials

[teal cedar]

polynomial long division?

[shrewd basin]

Yes

[teal cedar]

could you maybe show me what the division would look like

[shrewd basin]

gcd(a,b) = gcd(b,r) where a = bt + r

Hm I think there’s a latex package for that one sec

,polylongdiv

[teal cedar]

thats not what i thought when you said poly long division

[shrewd basin]

No that’s just Euclidean algorithm

In your case a is n^2 -2n b is 2n+1

[teal cedar]

okay... so let me try

should take just a minute or so

im struggling

[shrewd basin]

Chris do it

[teal cedar]

i got gcd(n^2 - 2n, 2n + 1) alright... and that should be equal to gcd(2n+1, r)... and n^2 - 2n = (2n + 1)t + r

should i just find r through euclidean algorithm?

[shrewd basin]

Honestly it’s just observation

You don’t need long division

[teal cedar]

but if i do that my r is not 5n^2

how did he get to 5n^2

[shrewd basin]

-2n(2n+1) gives you the right n coefficient

Then you only need to add 5n^2 > 0

[teal cedar]

what

how do you know that?

[shrewd basin]

What do you mean how do I know

-2n(2n+1) = -4n^2 -2n

and wwe need the -2n

[teal cedar]

okay? so you just add 5n^2 to get to n^2 - 2n

[shrewd basin]

Chris

Well yeah

[teal cedar]

im still not sure how i am supposed to think or get to this

[shrewd basin]

It’s like observation?

[teal cedar]

it could literally be anything

hello

[jolly parrot]

Willy the Explorer

[teal cedar]

well if its observation why did he say through long division?

[shrewd basin]

Im not sure have you tried doing long division

[teal cedar]

no... i dont even know what i should divide what for

i tried thinking but im not sure what to do

should i divide n^2 - 2n by 2n +1?

is that it?

but why?

[shrewd basin]

One sec I need to go to the lol

[teal cedar]

could you help me? 🥺

okay

,close

.close

[pearl pond]

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[shrewd basin]

It’s honestly easier to just do it by inspection

[jolly parrot]

Willy the Explorer

[teal cedar]

if i know what the result should be then yes

i closed because i figured it out lol

thanks guys

[shrewd basin]

What r u on about

[teal cedar]

the first part i figured out

actually i figured everything out by the end

so anyways, thanks you guys

[shrewd basin]

[jolly parrot]

Willy the Explorer

[shrewd basin]

calm stuff

[pearl pond]

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[shrewd basin]

💀

R u still playing lmao

Probably still suck at it

Yeah long day

Play Fortnite

[rustic gate]

[shrewd basin]

[late moon]

[rustic gate]

no i cant

yes its treason

i dont use overleaf if thats what you're thinking

[late moon]

Ok lint

[pearl pond]

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[ebon skiff]

if i know the derivitive of a function, can i find the function

[ebon skiff]

for example i know that the derivative is -5x^2+10x

[earnest stratus]

There are infinite functions though...

[ebon skiff]

yeah...

[earnest stratus]

Whose derivative is that.

[ebon skiff]

idk

[earnest stratus]

I do, I'm telling you.

[ebon skiff]

wait no

[earnest stratus]

What?

[ebon skiff]

ok tell me

[earnest stratus]

Tell you what?

[ebon skiff]

.

[earnest stratus]

I'm telling you that there's infinite functions with that derivative.

[ebon skiff]

oh

i thought you were about to tell me the function

[calm wing]

yes*
*to some extent

[earnest stratus]

yes, to some extent, now that's important.

You can find all those functions, after all there'll be only a difference in constant.

You just need to find the anti derivative.

Or integrate.

@ebon skiff

[ebon skiff]

oh

yeah thats right

so the indefinite integral of the derivative

[earnest stratus]

Yep.

Interestingly, you can find functions not just with the derivative.
This one that you mentioned, is one of the simplest cases, for what's called a differential equation.

[ebon skiff]

👍

thanks

.close

[pearl pond]

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[quick hare]

can someone help me solve this system of equations by elimination please?

[quick hare]

x + 3y = 9
2x + 6y = 30

this is the equation-

I've been trying for an hour now to solve it in different ways, and I've watched multiple videos

[earnest stratus]

Try again, this time show your work.

[quick hare]

I did -2(x + 3y = 9)
1(2x +6y = 30) and got 0x + 0y = 12

and then I tried doing
4(x + 3y =9)
-4(2x + 6y = 30) which then came out to X - 12y = 84 . Then I divided 12y = 84 by 12 and got y=-7 but I think that's also wrong

[fleet sky]

Double check your math on that top part. What would your y term be if you multiplied by -2?

[quick hare]

wouldn't it be -6?

[fleet sky]

yep

and what is -6y + 6y = ?

[quick hare]

0y

[fleet sky]

yep

you have 1y in your answer

so what you're left with is 0x + 0y = 12. what should that tell you about your system of linear equations?

[quick hare]

im not completely sure

[fleet sky]

it means we have no solutions

they're parallel line, they never cross

[quick hare]

ohhh ok

that makes a lot more sense

ok thank you

[fleet sky]

what really gives it away is that if you multiply the top equation by 2, you get 2x + 6y = 18.

In order for this to have a solution, the top line would have to equal the bottom line in all components (same x, same y, same constant). Otherwise, there's no solutions

no problem!

[quick hare]

I was having so much trouble with this and it turns out I only messed up one part . omg

Have a great rest of your day ^^

.close

[pearl pond]

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[polar steppe]

Am I misunderstanding something?

[plush bramble]

yes

,tex .maclaurin

[jolly parrot]

riemann

[plush bramble]

4th row

[polar steppe]

hmm

So if I had 1/ 1 - x I can write this as an the summation from 0 to infin (x^k)

hmmm

would it be 2x^n ?

[plush bramble]

you shouldn't skip this step

[polar steppe]

Okay

It's clear I'm not understanding what that's saying IMO.

I thought it was asking me to put the function I'm given in the form 1 / 1 - r

I guess the only difference is I still had the 2 in the denom

If not that, what I'm a missing

[pearl pond]

@polar steppe Has your question been resolved?

[polar steppe]

<@&286206848099549185>

[polar steppe]

.close

[pearl pond]

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[midnight haven]

Hey y'all

[midnight haven]

Can someone please help me with (b), the following is what I have at the moment. I think I need to use a z-chart to rewrite the fact that P(X < mu + r) = 0.96 but I am not sure how, considering the fact that the parameters of the normal distribution are (3, 2) and not (0, 1)

<@&286206848099549185>

[plush bramble]

If $X \sim N(\mu, \sigma^2)$, the $(X-\mu)/\sigma \sim Z$ where $Z \sim N(0,1)$

[jolly parrot]

riemann

[midnight haven]

First off, is it wrong for me to say that $X \sim N(3, 2)$ in this situation?

[jolly parrot]

Ω

[midnight haven]

Because that's what I wrote

[plush bramble]

depends on your convention

If your class's convention is to write standard deviation instead of variance in N(mean, stddev), then stick with it

[midnight haven]

Yeah, I think that's the case

[plush bramble]

no one is right, or wrong, just stick with whatever your class uses

[midnight haven]

Can you please explain what $(X-\mu)/\sigma \sim Z$ means

[jolly parrot]

Ω

[midnight haven]

I was thinking maybe I should use $x = \mu + z\sigma$

[jolly parrot]

Ω

[midnight haven]

But I'm not sure how to apply this

@plush bramble

[plush bramble]

that's algebraically the same as what i wrote

express this chain of inequalities in terms Z instead of X

[midnight haven]

Oh, did you mean $=$ and not $\sim$

[jolly parrot]

Ω

[midnight haven]

I got confused, my bad

[midnight haven]

I'm working on (b)?

[plush bramble]

oh my bad

[plush bramble]

same logic applies here

express the chain mu - r < X < mu + r in terms of Z

[midnight haven]

How’s this

[plush bramble]

,w 2 * CDF[NormalDistribution[0, 1], 1.75] - 1

[jolly parrot]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by rie.mann

[plush bramble]

looks right

[midnight haven]

Is it right to say that $P(X < \mu + r) = P(Z < r/2)$

[jolly parrot]

Ω

[plush bramble]

yes

[midnight haven]

Great

Thanks Riemann

.close

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[weary mango]

Are the help channels mainly for early university and below math?

[weary mango]

Sorry if this is not the correct channel to ask this

[fluid axle]

mostly yes

@weary mango

[weary mango]

I'm guessing that for advanced math help I go to the advanced math channels

[fluid axle]

yea you'll have more luck down there

[weary mango]

Got it

.close

[pearl pond]

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[burnt ocean]

can someone help

pls

[zealous shore]

These are probabilities, so they must add to 1. So we can set up x+2x^2+x(x+1)=1 and solve for x

[burnt ocean]

how do i do that

[zealous shore]

I'd start by multiplying the x(x+1) so you end up with x+2x^2+x^2+x=1 and then combine like terms

[burnt ocean]

4x^3 = 1 ?

what do i do now?

[zealous shore]

We can't combine the x's like that. When you have different powers of x, they have to stay separate. But the x^2 can add with the x^2 and the x can add with the x. So 2x^2+x^2 = 3x^2 and x+x=2x. So we end up with 3x^2+2x=1

(just realized I made a typo on my message before last, I dropped off a squared on one of the terms, it's fixed now)

[burnt ocean]

oh

after we get 3x^2+2x=1

what do we do to find x

[zealous shore]

We subtract 1 from each side to get 3x^2+2x-1=0. From here, we solve it like a quadratic, did you learn how to do that?

[burnt ocean]

i think i forgot it icl

[zealous shore]

This one can be tricky to factor. The quadratic formula is probably the safe way (it's a bit complicated to type, you might want to google it)

I can show how to factor it too, but it's just also tricky to explain lol

[burnt ocean]

bruhh

[zealous shore]

Was this taught a certain way in your class? I was trying to keep it simple since it seems like you're in a stats class and this is kind of an algebra problem

[burnt ocean]

we were taught it

but it was a while ago

ngl its 11pm and i got other work to do so i might drop this since its taking me so long

[zealous shore]

No worries, up to you on what you want to do. The short version is you factor it to (3x-1)(x+1)=0, separate it out to be 3x-1=0 and x+1=0, you get x=1/3 and -1, but -1 doesn't make sense so you take 1/3. Happy to explain how to get there but that's what the answer should be if you come back to it later

[burnt ocean]

1/3?

but

if i put it in as x

they wont add up to 100%

[zealous shore]

It does, 1/3 + 2*(1/3)^2 + 1/3(1+1/3) = 1

[pearl pond]

@burnt ocean Has your question been resolved?

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[viral sundial]

Need some help determining the results of a study.

[pearl pond]

@viral sundial Has your question been resolved?

[pearl pond]

@viral sundial Has your question been resolved?

[pearl pond]

@viral sundial Has your question been resolved?

[viral sundial]

<@&286206848099549185>

Even just having an idea of what im being asked to write helps. Im not sure if I'm supposed to calculate anything specific

[viral sundial]

.close

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[sly dagger]

prove by induction that the product of n even integers is even for n>=2

[sly dagger]

idk how to start this question

.close

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[vocal dove]

Why is this just not 60 * 59 * 58 ** * 2 **

i think the answer is C lol

60 options for the very first number

59 options for the second number

58 option for the third number

2 options for the direction sequence

so E (imo)

[quaint prawn]

Why would you need the *2?

And you're saying C but C doesn't have an *2

[vestal tapir]

they say it's E

[quaint prawn]

So you think it's C but you also think it's E

[vestal tapir]

at the end

[quaint prawn]

Yes but they say it twice so it's unclear

[vocal dove]

it's E

(my answer)

[vestal tapir]

yeah

[vocal dove]

sorry

C is the actual answer

E is what i think the answer should be

[vestal tapir]

they wrong

[vocal dove]

Sorry i keep mistyping

Let me rewrite lmao

The actual answer to the question is: C

However, according to me the answer should be E.

Why is this just not 60 * 59 * 58 * 2 ?

[vocal dove]

i'm pretty sure the answer in that site is correct

[vestal tapir]

i don't see any way to argue for C

[vocal dove]

i found this

https://gmatclub.com/forum/a-combination-lock-is-set-to-open-when-the-correct-direction-sequence-402442.html

Wasn't convinced tbh

Am i dumb lol

If you understand, can you explain it to me

[vestal tapir]

it would be like, you can't freely choose right-left-right

[vocal dove]

but it doesn't make sense to me

they say there's only one option

but there's 2 ways?

Lmao

in that case there's 1 way to choose the correct password too

not 60 * 59 * 58

[vestal tapir]

yeah they all don't get it, it's def E

[vocal dove]

LOL

[vocal dove]

wait what?

So you think it's C?

I need the 2 for 2 different possibilities for the direction sequence

[quaint prawn]

I'm not confident about that *2

[vestal tapir]

if you imagine that you can't freely rotate any combination, like you can't cross the starting position, you can argue for (60 * 59 * 58)(4/6)

[quaint prawn]

By that way, you could arrange that in 3P2 = 6 ways too

So *6

initially

60 (first slot) * 59 (second slot) * 58 (third slot)

Then you could have 60 58 59
58 59 60
58 60 59
59 60 50
59 58 60

[vocal dove]

that part is okay

the direction sequence is the problem here

[vestal tapir]

that's not how numbers work kavin

[quaint prawn]

no, what I'm saying is that

when we're multiplying 60 59 and 58

The logic on the basis of which we're doing it is that
In the first slot we're free to choose any of the 60 terms

In the second one, since we can't repeat, we can choose any of the 59

And in the third one, since we can't repeat again, we have to choose any of the remaining 58

Reduce the scale
Consider you have numbers 9, 5 and 6
You have to arrange them in groups of 2

So you'll do 3*2 = 6. But since you're doing an *2 for the sequence there, you're doing here too, so you have 13 permutations. What are the 12 permutations

[vocal dove]

@quaint prawn we get that though

lol

there's a direction sequence to consider as well

the problem is not with the 60P2 part

that's trivial

[quaint prawn]

Okay, explain it to me

[vocal dove]

What

[quaint prawn]

Explain why you did the *2. What exactly do you mean by direction?

[vocal dove]

"and" "correct three numbers from 1 to 60"

you only did the green part

[vestal tapir]

you can rotate the dial to the first number in two ways

[vocal dove]

where's the "either left-right-left or right-left-right"

[vestal tapir]

then you don;t have any choice

[vocal dove]

so the green part yes it's 60P2 like you said but the LrL or RLR is also another 2 possibilities

so 60P2 * 2 according to me tbh

it should not be 60P2 (which happens to be the answer)

[vocal dove]

do i just assume like all of them have lost the plot lol

and we're the only ones that are right

that seems highly unlikely 😭

[vestal tapir]

there aren't that many people

[vocal dove]

but i can't wrap my head around this tbh

[vocal dove]

i've seen in another chinese site

the answer is the same

but they write in chinese

[vestal tapir]

i think they interpret it weirdly, like both directions are actually allowed

so they aren't a part of the combination or anything

[vocal dove]

yeah i think so too

is there something special about combination locks?

maye that's what we're missing?

i'll find you the chinese solution

If it makes sense to you

[vestal tapir]

no i tried to think of a lock that doesn;t let you cross the initial state

then it's × 4/6 instead of × 2

[vocal dove]

lol bruh

Here, it's translated btw

[vestal tapir]

then e.g. 10 15 80 is not allowed

well yeah they explicitly say the direction is not part of the combination

[vocal dove]

wait what lmao

honestly i read the solution just with you

"both methods can be unlocked"

[vestal tapir]

"it's interference information"

[vocal dove]

??????

But the people in that gmat club said otherwise

I have no clue what they were waffling about tbh

Lmao

The chinese site makes sense but i'm not sure how i was supposed to understand this on an actual exam

[vestal tapir]

they both say the same thing, "i don't care about direction stuff, ignore that, because reasons"

[vocal dove]

so 10 15 80 works both LRL or RLR

ig?

[vestal tapir]

this would happen if they are justifing the answer which they already know is C

[vocal dove]

They're the "experts" apparently

Guess i have to learn english

Cuz 100% in an exam setting i'm choosing E without a second though

and not pondering over the grammatical structure of a math question lmfao

[vestal tapir]

E is right

[vocal dove]

Okay fair enough

Actually

i'm thinking maybe it has to do with combinations? i.e combination lock

[vestal tapir]

10 15 80 can't be entered with this hypothetical lock

[vocal dove]

AB and BA are the same thing boom

so only 1 possibility

maybe that's how they derived the name

123 and 321 and all the other degenerates are the same so only 1?

[vestal tapir]

if thre was a way to justify C I would already came up with it

[vocal dove]

😭

okay i'm with you

btw could you help me with something trivial

What does this mean? The symmetrical set part

where is the symmetry?

basically i was checking some descriptive stats and i saw mean = median = mode in a symmetrical distribution (but that's continuous)

was looking at an unordered set (discretized one)

this should be trivial though i'm not sure where i see the symmetry lmao

[vestal tapir]

no idea sorry

[vocal dove]

owh

okay np thanks

[quaint prawn]

Okay. Consider a set of 4 numbers arranged into a group of 2. Consider the same condition with the direction sequence. Tell me what impact it makes

I'm still not clear

[vestal tapir]

given any choice of 3 numbers (including order), they can be entered in two ways

illustrated

you approach the first number either way, then the second number you have to go the opposite direction, same for the third

[quaint prawn]

Okay, so clockwise could be 1 3 5 and anticlockwise could be 1 5 3?

[vestal tapir]

the lock can tell which one you did, and only open on one of them

[quaint prawn]

Give me number combinations for clockwise and anticlockwise

[vestal tapir]

no, both would be the saem sequence

e,g, 5 1 3 and 5 1 3

[quaint prawn]

I partially get the hang of it but I'm not completely convinced, but alright

[vestal tapir]

if the lock resets the sequence when you hit 0 as you could imagine it does, then you can't enter 1 3 5 at all
then for each of 60×59×58/6 sets you can enter them in 4 ways

so × 4/6 which is not an option, so it's not like there's a tricky thing that locks do we're missing

[pearl pond]

@vocal dove Has your question been resolved?

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[pearl pond]

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[meager wharf]

What is the difference between these 2?

[regal herald]

what do you mean?

[ebon skiff]

"then find" is different

And downwards its different

[meager wharf]

I don’t know how to solve the problems and what each part means

Like I don’t know how to identify what transformation it is

[pearl pond]

@meager wharf Has your question been resolved?

[pearl pond]

Channel closed

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[shadow cargo]

So,y=-6 but i find that y=1 what am i doing wrong

[shadow cargo]

Let me check

Ok checked,its correct

Why is it a minus there though

[viscid sierra]

$\frac{-12\cdot2\left(y-2\right)}{4}=-6\left(y-2\right)=-6y+12$

[jolly parrot]

B-eard

[shadow cargo]

Oooh

Ok thanks

.close

[pearl pond]

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Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[summer torrent]

what is the most simple way to prove that 5/2 ^ 5/2 < 10

[tropic saddle]

square both sides

rearrange

[summer torrent]

can you send example please?

[shadow cargo]

You solve 5/2(2.5)

And its 2.5^2.5

Then you use a calculator

[hollow cobalt]

Proof by a calculator

[summer torrent]

i cant use calculator on test

[hollow cobalt]

Seriously though I would square both sides definitely

[summer torrent]

square with what

[hollow cobalt]

[shadow cargo]

Then solve it

Its not a large number

[hollow cobalt]

0 < a < b means that a^2 < b^2

I am talking about this

[summer torrent]

bro i need to prove that 5/2 ^ 5/2 <10

i cant just solve it

[hollow cobalt]

I know

[fast zealot]

thats why you square both sides :D

[summer torrent]

with what

[hasty eagle]

https://tenor.com/view/theres-something-very-familiar-about-all-this-deja-vu-i-feel-like-ive-done-this-before-this-feels-familiar-i-feel-live-ive-been-here-before-gif-16489734

[shrewd basin]

What you mean with what

[hollow cobalt]

Wdym by "with what"?

[fast zealot]

raise to the power of 2

is that better?

[summer torrent]

send picture of what it would look like please english is not my first languagne

[shadow cargo]

Solve 5/2^5/2 and see if its lower than 10

[fast zealot]

I see

[shrewd basin]

What r u talking about solving

[hasty eagle]

$a*a = a^2$

that

[jolly parrot]

TooManyCooks

[summer torrent]

you need calculator for that

[shadow cargo]

Not really

[hollow cobalt]

[ \left( \frac52 \right)^{\frac52} < 10 ]
[ \left( \left( \frac52 \right)^{\frac52} \right)^2 < 10^2 ]

[jolly parrot]

A Lonely Bean

[summer torrent]

what are you even doing here you are not helping

[shadow cargo]

Ok sorry i am leaving then

[summer torrent]

how does that prove it